Given two plane isometries $g\ , f$ and $f^{'} = g\circ f \circ g^{-1}$ prove that:
- If $P$ is a fixed point of $f$ then $g\left(P\right)$ is a fixed point of $f^{'}$ and if $Q$ is a fixed point of $f^{'}$ then then $g^{-1}\left(Q\right)$ is a fixed point of $f$.
- Prove that $f$ and $f^{'}$ are isometries of the same type (e.g if $f$ is a rotation isometry then so is $f^{'}$)
My attempt: I've tried evaluating $f^{'}$ at $A$ and substituting it for the explicit expression but with no luck.
I've proven that $f^{'}$ is an isometry and that topological orientation is preserved iff $f$ preserves topological orientation.
And I know that if some isometry $h$ preserves topological orientation it has to be one of the following:
- Identity
- Rotation
- Translation
And that if $h$ has a fixed point it has to be :
- Identity
- Rotation
But I could not quite formulate a proof using all this data.
NOTE: This is my first university level course and I'm 14 so I don't have much experience neither with mathematics related resources over the net or with rigorous proofs, so sorry if something is not according to SE policy.
Hints (in order of the questions asked):
For $P$ a fixed point of $f$, $g(f(g^{-1}(g(P))))=$?
Assuming $Q$ is a fixed point of $f'$ amounts to saying that $Q=g(f(g^{-1}(Q)))$. Apply $g^{-1}$ to both sides and examine what you have.
$g(Id(g^{-1}(x)))=$? (The $Id$ denotes the identity transformation.)
The very first pair of questions help you decide that $f$ and $f'$ either both have a fixed point, or both do not have a fixed point. This will help you decide the statement for rotations and translations.